新规范箱涵结构设计7月

1、1、孔径及净空约定:净跨径L 0 = 3.0m 1、轴向力以杆件受压为正。净高h 0 = 2.5m2、弯矩以使箱涵内框一侧受拉为正。2、设计安全等级三级3、截面剪力:使计算截面逆时针转动为正。结构重要性系数r 0 =0.9前提条件:1支座在杆件下方;3、汽车荷载2x为计算截面距左端支座之距离。荷载等级公路 级4、支座反力以受压为正。4、填土情况5、以梁为隔离体,固端梁简化为简支梁时,涵顶填土高度H =0.8m 图中所加的杆端力(轴力、弯矩及支反力均为正值,土的内摩擦角 =30°若计算得到的值为负值,则表示实际方向与图中所示填土容重1 =18kN/m 3的方向相反。地基容许承载力0 =

2、200kPa6、本表格默认箱涵内水深为满水,也可以修改表格,5、建筑材料先不考虑水压力得到涵洞基底应力max,普通钢筋种类HRB335最后要求的基底应力为:max+h 水*10KPa。主钢筋直径12mm 钢筋弹性模量Es =200000MPa 钢筋抗拉强度设计值f sd =280MPa7、本表格未考虑砼收缩、温度变化的影响。涵身混凝土强度等级C 208、本表格已考虑涵洞设计细则中竖向土压力系数的影响。涵身混凝土抗压强度设计值f cd =9.2MPa 9、本表格已考虑偏心受压构件裂缝宽度验算。涵身混凝土抗拉强度设计值f td = 1.06MPa 钢筋混凝土重力密度2 =25.0kN/m 3基础混

3、凝土强度等级C 15混凝土重力密度3 =24.0kN/m 3(一截面尺寸拟定 (见图L-01顶板、底板厚度 =0.3m C 1 =0.3m 侧墙厚度t =0.28m C 2 =0.5m 横梁计算跨径L P = L 0+t = 3.28m L = L 0+2t = 3.56m 侧墙计算高度h P = h 0+ = 2.8m h = h 0+2 =3.1m 基础襟边 c =0.2m 基础高度 d =0.4m 基础宽度B = 3.96mJTG/T D65-04-2007表9.2.2h/D =0.22竖向土压力系数Ks =1.09图 L-01(二荷载计算1、恒载恒载竖向压力p 恒 = Ks 1H+2

4、=23.19kN/m 2恒载水平压力钢 筋 混 凝 土 箱 涵 结 构 设 计一 、 设 计 资 料 二 、 设 计 计 算顶板处e P1 = 1Htan 2(45°-/2 = 4.80kN/m 2底板处e P2 = 1(H+htan 2(45°-/2 =23.40kN/m 22、活载汽车后轮着地宽度0.6m,由公路桥涵设计通用规范(JTG D602004第4.3.4条规定,按30°角向下分布。汽车后轮横向分布宽(此处计算 横向最多只考虑两辆汽车-共4个车轮荷载叠加的情况一个后轮单边荷载> 1.3/2 m横向分布宽度(m< 1.8/2 m (按30度扩

5、散,涵9.2.3条故横向分布宽度a = (0.6/2+Htan30°³2+1.3 =2.824m考虑横向车轮数目n横=2同理,纵向,汽车后轮着地长度0.2m一个车轮单边的纵向分布宽度0.2/2+Htan30°=0.562 m < 1.4/2 m 故 b = (0.2/2+Htan30°³2 =1.124m 考虑纵向车轮数目n纵=1考虑的总车轮重量G =140kN 车辆荷载垂直压力q 车 = G /(a³b =44.12kN/m 2车辆荷载水平压力e 车 = q 车tan 2(45°-/2 =14.71kN/m 2(三内

6、力计算1、构件刚度比K = (I 1/I 2³(h P /L P =1.05 2、节点弯矩和轴向力计算(1a种荷载作用下 (图L-02涵洞四角节点弯矩M aA = M aB = M aC = M aD =-1/(K+1²pL P 2/12横梁内法向力N a1 = N a2 =0侧墙内法向力N a3 = N a4 =pL P /2恒载p = p 恒 =23.19kN/m 2M aA = M aB = M aC = M aD =-10.14kN ²m N a3 = N a4 =38.04kN 车辆荷载p = q 车 =44.12kN/m 2M aA = M aB =

7、M aC = M aD =-19.30kN ²mN a3 = N a4 =72.36kN (2b种荷载作用下 (图L-03M bA = M bB = M bC = M bD =-K/(K+1²ph P 2/12N b1 = N b2 =ph P /2N b3 = N b4 =0恒载p = e P1 =4.80kN/m 2M bA = M bB = M bC = M bD =-1.61kN ²m N b1 = N b2 =6.72kN(3c 种荷载作用下 (图L-04M cA = M cD =M cB = M cC =N c1 =ph P /6+(M cA -M c

8、B /h P N c2 =ph P /3-(M cA -M cB /h PN c3 = N c4 =0恒载p = e P2-e P1 =18.60kN/m 20.76 m0.6/2+Htan30°= M cA = M cD =-3.43kN ²m M cB = M cC =-2.80kN ²m N c1 =8.45kN N c2 =17.59kN(4d 种荷载作用下 (图L-05图 L-04M dA =-K(K+3/6(K 2+4K+3+(10K+2/(15K+5²ph P 2/4M dB =-K(K+3/6(K 2+4K+3-(5K+3/(15K+5&

9、#178;ph P 2/4M dC =-K(K+3/6(K 2+4K+3+(5K+3/(15K+5²ph P 2/4M dD =-K(K+3/6(K 2+4K+3-(10K+2/(15K+5²ph P 2/4N d1 =(M dD -M dC /h P N d2 =ph P -(M dD -M dC /h PN d3 = -N d4 =-(M dB -M dC /L P 车辆荷载p = e 车 =14.71kN/m 2M dA =-19.82kN ²m M dB =9.00kN ²m M dC =-13.92kN ²m M dD =14.90k

10、N ²m N d1 =10.29kN N d2 =30.88kNN d3 = -N d4 =-6.99kN(5节点弯矩、轴力计算及荷载效应组合汇总表按公路桥涵设计通用规范(JTG D602004第4.1.6条进行承载能力极限状态效应及第4.1.7条正常使用极限状态效应组合 (3左侧墙 (图L-08 (4右侧墙 (图L-09 M d =71.70 kN²m ,N d =35.66 kN, V d =9.78 kN l0/h=10.93<=14计算L0 =3.28 m ,e0 = M d/N d = 2.011mi =h/121/2 =0.087m长细比L0/i =37.

11、87> 17.5由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.10条1 = 0.2+2.7e0/h0 =21.081> 1.0 ,取1 =1.002 = 1.15-0.01l0/h = 1.041> 1.0 ,取2 =1.00 =1+(l0/h212h0/1400e0 = 1.011由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e0+h/2-a = 2.143mr0N d e =f cd bx(h0-x/2 由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.10条0.50

12、9;10-32f td bh0 =137.8 kN>r0V d =8.8 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D622004第9.3.13条构造要求配置箍筋。(2结点l0 =3.28 m ,h = +C1 =0.60 m , a =0.04 m ,h0 =0.56 m , b =1.00 m , M d =64.84 kN²m ,N d =35.66 kN, V d =156.73 kN计算L0 =3.28 m ,e0 = M d/N d = 1.818mi =h/121/2 =0.173m长细比L0/i =18.94> 17.5由公路钢筋混凝土及预应力

13、混凝土桥涵设计规范(JTG D622004第5.3.10条1 = 0.2+2.7e0/h0 =8.967> 1.0 ,取1 =1.002 = 1.15-0.01l0/h = 1.095> 1.0 ,取2 =1.00 =1+(l0/h212h0/1400e0 = 1.007由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e0+h/2-a = 2.090mr0N d e =f cd bx(h0-x/267.08 =9200x(0.56-x/2解得x =0.013 mb h0 =0.56³0.56 =0.314 m故为大偏心受压构件。依D

14、62-2004规范公式5.3.5-1若为小偏心受压构件,则需要调整钢筋As的计算公式A s = (f cd bx-r0N d/f sd =0.0003183m2= 318.3mm2 = 100A s/(bh0 =0.06 %<0.2 %应按最小配筋率配置受拉钢筋。选用 12 100 mm,实际 A s =1131.0mm20.51³10-3f cu,k1/2bh0 =1277.2 kN>r0V d =141.1 kN故抗剪截面符合公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.9条的要求。由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D62

15、2004第5.2.10条0.50³10-32f td bh0 =296.8 kN>r0V d =141.1 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D622004第9.3.13条构造要求配置箍筋。2、底板 (A-D钢筋按左、右对称,用最不利荷载计算。(1跨中l0 =3.28 m ,h =0.30 m , a =0.04 m ,h0 =0.26 m , b =1.00 m , M d =70.82 kN²m ,N d =77.26 kN, V d =27.13 kN l0/h=10.93<=14计算L0 =3.28 m ,e0 = M d/N d =

16、0.917mi =h/121/2 =0.087m长细比L0/i =37.87> 17.5由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.10条1 = 0.2+2.7e0/h0 =9.718> 1.0 ,取1 =1.002 = 1.15-0.01l0/h = 1.041> 1.0 ,取2 =1.00 =1+(l0/h212h0/1400e0 = 1.024由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e0+h/2-a = 1.049mr0N d e =f cd bx(h0-x/272.93 =9200x(0.

17、26-x/2解得x =0.033 mb h0 =0.56³0.26 =0.146 m故为大偏心受压构件。依D62-2004规范公式5.3.5-1若为小偏心受压构件,则需要调整钢筋As的计算公式A s = (f cd bx-r0N d/f sd =0.0008202m2= 820.2mm2 = 100A s/(bh0 =0.32 %>0.2 %符合公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第9.1.12条的要求。选用 12 130 mm,实际 A s =870.0mm20.51³10-3f cu,k1/2bh0 =593.0 kN>r0V d

18、 =24.4 kN故抗剪截面符合公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.9条的要求。由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.10条0.50³10-32f td bh0 =137.8 kN>r0V d =24.4 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D622004第9.3.13条构造要求配置箍筋。(2结点l0 =3.28 m ,h = +C1 =0.60 m , a =0.04 m ,h0 =0.56 m , b =1.00 m , M d =73.99 kN²m ,N d =77.

19、26 kN, V d =137.16 kN计算L0 =3.28 m ,e0 = M d/N d =0.958mi =h/121/2 =0.173m长细比L0/i =18.94> 17.5由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.10条1 = 0.2+2.7e0/h0 = 4.817> 1.0 ,取1 =1.002 = 1.15-0.01l0/h = 1.095> 1.0 ,取2 =1.00 =1+(l0/h212h0/1400e0 = 1.012由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e0+h/

20、2-a = 1.230mr0N d e =f cd bx(h0-x/285.50 =9200x(0.56-x/2解得x =0.017 mb h0 =0.56³0.56 =0.314 m故为大偏心受压构件。依D62-2004规范公式5.3.5-1若为小偏心受压构件,则需要调整钢筋As的计算公式A s = (f cd bx-r 0N d /f sd =0.0003053m 2= 305.3mm 2 = 100A s /(bh 0 =0.05 %<0.2 %应按最小配筋率配置受拉钢筋。100 mm,实际 A s =1131.0mm 20.51³10-3f cu,k 1/2b

21、h 0 =1277.2 kN>r 0V d =123.4 kN故抗剪截面符合公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.9条的要求。由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.10条0.50³10-32f td bh 0 =296.8 kN>r 0V d =123.4 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D622004第9.3.13条构造要求配置箍筋。3、左、右侧板 (B-A,C-D(1板中l 0 =2.80 m ,h =0.28 m , a =0.04 m ,h 0 =0.24 m ,b =

22、1.00 m ,M d =25.76 kN²m ,N d =156.73 kN , V d =17.14 kNl 0/h=10.00<=14计算L 0 =2.80 m ,e 0 = M d /N d =0.164m i =h/121/2 =0.081m 长细比L 0/i =34.64> 17.5由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.10条1 = 0.2+2.7e 0/h 0 =2.049> 1.0 ,取1 =1.002 = 1.15-0.01l 0/h =1.050> 1.0 ,取2 =1.00 =1+(l 0/h212h

23、 0/1400e 0 =1.104由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e 0+h/2-a =0.281mr 0N d e =f cd bx(h 0-x/239.71 =9200x(0.24-x/2解得x =0.019 mb h 0 =0.56³0.24 =0.134 m若为小偏心受压构件,则需要调整钢筋As的计算公式A s = (f cd bx-r 0N d /f sd =0.0001111m 2= 111.1mm 2 = 100A s /(bh 0 =0.05 %<0.2 %应按最小配筋率配置受拉钢筋。230 mm,实际 A

24、 s =491.7mm 20.51³10-3f cu,k 1/2bh 0 =547.4 kN>r 0V d =15.4 kN故抗剪截面符合公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.9条的要求。由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.2.10条0.50³10-32f td bh 0 =127.2 kN>r 0V d =15.4 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D622004第9.3.13条构造要求配置箍筋。(2结点l 0 =2.80 m ,h = t+C 2 =0.78 m ,a

25、 =0.04 m ,h 0 =0.74 m ,b =1.00 m ,M d =73.99 kN²m ,N d =137.16 kN , V d =77.26 kN 计算L 0 =2.80 m ,e 0 = M d /N d =0.539m i =h/121/2 =0.225m 长细比L 0/i =12.44< 17.5由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.10条,不考虑偏心距增大系数。1 = 0.2+2.7e 0/h 0 =2.168> 1.0 ,取1 =1.002 = 1.15-0.01l 0/h =1.114> 1.0 ,取

26、2 =1.00 =1+(l 0/h212h 0/1400e 0选用 12 选用 12 =1.000由公路钢筋混凝土及预应力混凝土桥涵设计规范(JTG D622004第5.3.5条e = e 0+h/2-a =0.889mr 0N d e =f cd bx(h 0-x/2109.79 =9200x(0.74-x/2解得x =0.016 mb h 0 =0.56³0.74 =0.414 m若为小偏心受压构件,则需要调整钢筋As的计算公式A s = (f cd bx-r 0N d /f sd =9.492E-05m 2= 94.9mm 2 = 100A s /(bh 0 =0.01 %<0.2 %应按最小配筋率配置受拉钢筋。70 mm,实际 A s =1615.7mm 20.51³10-3f cu,k 1/2bh 0 =1687.8 kN>r 0V d =69.5 kN故抗剪截面